pub fn new_birthday_probability(n: u32) -> f64 {
    assert!(
        n > 2,
        "人数必须大于2"
    );
    
    if n > 365 {
        return 1.0;
    }
    // 至少两个人在同一天过生日的概率 = 1 - 所有人生日都不同的概率
    const DAYS_OF_YEAR: f64 = 365.0;
    let mut prob = 1.0; // 所有人生日不同
    for i in 0..n {
        prob *= (DAYS_OF_YEAR - i as f64) / DAYS_OF_YEAR;
    }
    let ans  = 1.0 - prob;

    (ans * 10000.0).round() / 10000.0
}
